A

Problem description

"Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.

A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space.

The second line contains n space-separated integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 100), where a_i is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: a_i ≥ a_i + 1).

Output

Output the number of participants who advance to the next round.

Examples

Input

8 5
10 9 8 7 7 7 5 5

Output

6

Input

4 2
0 0 0 0

Output

0

Note

In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.

In the second example nobody got a positive score.

解题思路：题目比较简单，就是找出n个人里的得分不小于第k个人的得分的人数，并且要求这个分数是正数，不能为0。

AC代码：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,k,num=0,a[105];
    cin>>n>>k;
    for(int i=0;i<n;++i)cin>>a[i];
    for(int i=0;i<n;++i)
        if(a[i]!=0 && a[i]>=a[k-1])num++;
    cout<<num<<endl;

    return 0;
}